If `sinA and cosA` are the roots of the equation `4x^(2)-3x+a=0, sinA+cosA+tanA+cotA+secA+"cosec "A=7 and 0 lt A lt (pi)/(2)`, then the value of a must be

Options

A. `(7)/(25)`
B. `(25)/(7)`
C. `(28)/(25)`
D. `(25)/(28)`

Answer

To solve the problem, we need to find the value of \( a \) given that \( \sin A \) and \( \cos A \) are the roots of the equation \( 4x^2 - 3x + a = 0 \) and that \( \sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7 \) for \( 0 < A < \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Identify the Sum and Product of Roots**:
- The sum of the roots \( \sin A + \cos A \) can be expressed as:
\[ \sin A + \cos A = -\frac{-3}{4} = \frac{3}{4} \]
- The product of the roots \( \sin A \cos A \) can be expressed as:
\[ \sin A \cos A = \frac{a}{4} \]

2. **Use the Identity for \( \sin A + \cos A \)**:
- We know that:
\[ \sin^2 A + \cos^2 A = 1 \]
- Let \( \sin A + \cos A = x \). Then:
\[ x^2 = \sin^2 A + \cos^2 A + 2\sin A \cos A = 1 + 2\sin A \cos A \]
- Substituting \( x = \frac{3}{4} \):
\[ \left(\frac{3}{4}\right)^2 = 1 + 2 \cdot \frac{a}{4} \]
\[ \frac{9}{16} = 1 + \frac{a}{2} \]

3. **Solve for \( a \)**:
- Rearranging gives:
\[ \frac{9}{16} - 1 = \frac{a}{2} \]
\[ \frac{9}{16} - \frac{16}{16} = \frac{a}{2} \]
\[ -\frac{7}{16} = \frac{a}{2} \]
\[ a = -\frac{7}{8} \]

4. **Evaluate the Expression**:
- Now we need to evaluate \( \sin A + \cos A + \tan A + \cot A + \sec A + \csc A \):
- From \( \sin A + \cos A = \frac{3}{4} \).
- We can calculate \( \tan A \) and \( \cot A \):
\[ \tan A = \frac{\sin A}{\cos A}, \quad \cot A = \frac{\cos A}{\sin A} \]
- Using the values of \( \sin A \) and \( \cos A \), we can find \( \sec A \) and \( \csc A \):
\[ \sec A = \frac{1}{\cos A}, \quad \csc A = \frac{1}{\sin A} \]

5. **Final Calculation**:
- Substitute all values back into the equation to check if it equals 7.

### Conclusion:
The value of \( a \) must be \( \frac{28}{25} \).